Commutant of *-algebra $A\subset B(\mathsf{H})$ is strongly closed

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A quick one.

Suppose that $(f_\lambda)\subset A'$ is strongly convergent to $f\in B(\mathsf{H})$. That is for all $x\in \mathsf{H}$:

$$f_\lambda(x)\to f(x).$$

Let $g\in A$. We want to show that $gf=fg$, as this implies that $f\in A'$. Let $g(x)\in\mathsf{H}$ so that:

$$\lim_\lambda f_\lambda(g(x))=f(g(x)).$$ Also, as $f_\lambda\in A'$, we have: $$\lim_\lambda f_\lambda(g(x))=\lim_{\lambda}g(f_\lambda(x)).$$

I presume that I am supposed to conclude that this equals $g(f(x))$... but how?

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You are applying a continuous function (the bounded operator $g$) to the convergent net $\{f_\lambda\}$.